Asked by Lost One
find the derivative of y =(x+3)^3 / sqrt(x^2 -1) using logarithmic differentation.
Answers
Answered by
Steve
y =(x+3)^3 / √(x^2 -1)
lny = 3ln(x+3) - 1/2 ln(x^2-1)
1/y y' = 3/(x+3) - x/(x^2-1)
1/y y' = (3(x^2-1) - x(x+3))/((x+3)(x^2-1))
1/y y' = (2x^2-3x-3)/((x+3)(x^2-1))
y' = (x+3)^3 / √(x^2 -1) * ((2x^2-3x-3)/((x+3)(x^2-1)))
= ((x+3)^2 * (2x^2-3x-3))/(x^2-1)^(3/2)
lny = 3ln(x+3) - 1/2 ln(x^2-1)
1/y y' = 3/(x+3) - x/(x^2-1)
1/y y' = (3(x^2-1) - x(x+3))/((x+3)(x^2-1))
1/y y' = (2x^2-3x-3)/((x+3)(x^2-1))
y' = (x+3)^3 / √(x^2 -1) * ((2x^2-3x-3)/((x+3)(x^2-1)))
= ((x+3)^2 * (2x^2-3x-3))/(x^2-1)^(3/2)
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